====== Proof of the Pythagorean Theorem ======

===== Pythagorean theorem =====

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In a right triangle the square of the hypotenuse is equal to the sum of the squares of the sides containing the right angle. 
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I've been puzzling over a proof for this for years, and it finally dawned on me. (Eureka!) It's all in how you draw it...

===== Proof #1 =====

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https://docs.google.com/drawings/d/18o9L3R3PQ1LKha5RaGQ8JI2bjzhNeCgWnX7cfKPfTjg/edit
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Given the triangle formed by \(a\), \(b\) (choosing \(b\geq a\)) and \(c\), we can construct a square with total area \(c^2\). As shown, we can fit four triangles, each with area \(a b/2\), into the large square, leaving an inner square with area \((b-a)^2\). Thus, the total area of the large square is
\[
\begin{array}{rl}
  c^2 & = 4 (a b/2) + (b-a)^2 \\
      & = 2 a b + a^2 + b^2 - 2 a b \\
      & = a^2 + b^2 .
\end{array}
\]

Hence, the Pythagorean theorem.

===== Proof #2 =====

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https://docs.google.com/drawings/d/1MjGAWhQzsbPG-3s9SWL9op6IAAVj_RI_eTa3SMJrwSo/edit
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I found another proof, which [[http://www.jimloy.com/ | Jim Loy]] (visit [[http://www.jimloy.com/geometry/pythag.htm | his Pythagorean page]]!) told me is due to Legendre. It relies on recognizing that you can subdivide a triangle forming two sub-triangles //similar// to each other and the original. (I won't prove this.) Then, from the figure above, and from the properties of similar triangles
\[
  \frac{a}{e} = \frac{c}{a} \text{ thus } a^2 = c e
\]
and
\[
  \frac{b}{f} = \frac{c}{b} \text{ thus } b^2 = c f.
\]
Adding the two results together gives
\[
\begin{array}{rl}
  a^2 + b^2 & = c e + c f \\
            & = c (e+f) \\
            & = c^2 .
\end{array}
\]

Hence, the Pythagorean theorem. 

 --- //[[rik.blok@ubc.ca|Rik Blok]] 1997//

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